The angular frequency of the damped oscillator is given by `omega=sqrt((k)/(m)-(r^(2))/(4m^(2)))` ,where k is the spring constant, m is the mass of the oscillator and r is the damping constant. If the ratio `r^(2)//(m k)` is 8% ,the change in the time period compared to the undamped oscillator

Options

A. increases by 1%
B. increases by 8%
C. decreased by 1%
D. decrease

Explanation

To solve the problem of finding the change in the time period of a damped oscillator compared to an undamped oscillator, we can follow these steps:

### Step 1: Understand the given formula for angular frequency
The angular frequency of a damped oscillator is given by:
\[ \omega = \sqrt{\frac{k}{m} - \frac{r^2}{4m^2}} \]
where \(k\) is the spring constant, \(m\) is the mass, and \(r\) is the damping constant.

### Step 2: Identify the undamped angular frequency
For an undamped oscillator, the angular frequency is:
\[ \omega_0 = \sqrt{\frac{k}{m}} \]

### Step 3: Relate the two angular frequencies
We can express the damped angular frequency in terms of the undamped angular frequency:
\[ \omega = \sqrt{\frac{k}{m}} \sqrt{1 - \frac{r^2}{4mk}} = \omega_0 \sqrt{1 - \frac{r^2}{4mk}} \]

### Step 4: Write the time period equations
The time period \(T\) of an oscillator is related to the angular frequency by:
\[ T = \frac{2\pi}{\omega} \]
For the undamped oscillator:
\[ T_0 = \frac{2\pi}{\omega_0} \]
For the damped oscillator:
\[ T = \frac{2\pi}{\omega} = \frac{2\pi}{\omega_0 \sqrt{1 - \frac{r^2}{4mk}}} \]

### Step 5: Find the ratio of the time periods
The ratio of the time periods is given by:
\[ \frac{T}{T_0} = \frac{1}{\sqrt{1 - \frac{r^2}{4mk}}} \]

### Step 6: Use the given ratio
We are given that:
\[ \frac{r^2}{mk} = 0.08 \quad \text{(8%)} \]
Thus,
\[ \frac{r^2}{4mk} = \frac{0.08}{4} = 0.02 \]

### Step 7: Substitute into the time period ratio
Now substituting this into the ratio of the time periods:
\[ \frac{T}{T_0} = \frac{1}{\sqrt{1 - 0.02}} = \frac{1}{\sqrt{0.98}} \]

### Step 8: Calculate the change in time period
Using the approximation for small changes, we can expand \(\sqrt{0.98}\):
\[ \sqrt{0.98} \approx 1 - \frac{0.02}{2} = 1 - 0.01 = 0.99 \]
Thus,
\[ \frac{T}{T_0} \approx 1.01 \]
This indicates that:
\[ T \approx 1.01 T_0 \]

### Step 9: Determine the percentage change
The percentage change in the time period is:
\[ \text{Change} = \left(\frac{T - T_0}{T_0}\right) \times 100\% = (1.01 - 1) \times 100\% = 1\% \]

### Conclusion
The change in the time period compared to the undamped oscillator is approximately **1% increase**.

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